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# 9.1: Application of the Rules to Complex Sentences

This is going to be a short chapter. You really have all the facts about truth trees. It only remains to see how to apply these facts in some new ways.

In the last chapter I was careful to give you only problems in which the sentences were very simple. But now that you have the hang of the rules, let's see how to apply them in testing the validity of an argument like this: Following the suggestion of working first on nonbranching lines, I began with line 2. But what, then, should I do with line l? Line 1 is a disjunction of conjunctions. Which rule applies? And how? Keep in mind what the rules are supposed to do. Our objective is to make line 1 true, in all the minimally sufficient ways that this can be done. Because line 1 is a disjunction, we must do this by making each of the disjuncts true along a separate leg of a branch. That is, we must make the full subsentence 'A&B' true along one branch and the full subsentence '~A&C' true along a second branch. The subsentences 'A&B' and '~A&C' are themselves compound sentences to which we must apply the rule for conjunction, which I have done in lines 6 and 7.

How can you tell which rule applies to line l? Ask yourself: What do 1 have to do to the whole sentence appearing on line 1 which will guarantee that the whole sentence is true? The answer will direct you to the components which must be written at points farther down on the tree.
To see more clearly how this works, let us look at some more examples. If the sentence is

(A&B)⊃(CvD)

I say to myself: This sentence is a conditional, the antecedent of which is the conjunction 'A&B' and the consequent of which is the disjunction 'CvD'. A conditional can be made true by making the antecedent false. And it can alternatively be made true by making the consequent true. So at the bottom of every open path on which '(A&B)⊃(CvD)' appears, I must write the branch What should I do with

-[(BvC)⊃A]?

This sentence is a negated conditional. To make it true I must simultaneously make the conditional's antecedent true and its consequent false. So at the bottom of every open path on which it appears, I must write the stack

BvC
~A

We will look at some nastier examples in a moment. But first let's discuss an explicit prescription for applying the rules to sentences no matter how complex. Suppose you are faced with a very complex sentence and you are not sure which rule applies. Ask yourself the following question: In building this sentence up from smaller parts, what was the last step? What was the very last connective that got used in putting together this compound sentence from its components?

I hope that by this time you are recognizing the idea of a sentence's main connective from chapter 1. (If you did the chapters on natural deduction, you may find this section repetitious. Please bear with those who are learning the tree method without having learned natural deduction first.)

The Main Connective in a compound sentence is the connective which was used last in building up the sentence from its component or components.

To determine which rule applies to a sentence, first determine the sentence's main connective. If the main connective is not '~' all you have to do is to apply the rule for the main connective. If the main connective is '~', determine the main connective of the negated sentence. Then apply the corresponding rule ~v, ~&, ~⊃, ~≡, or ~~. Let us see how the rules apply to a few more examples. Consider

(A⊃B)⊃[(CvA)⊃B)]

What is the main connective? It is '⊃'. But which occurrence of '⊃'? It's the second occurrence. The parentheses tell you that the very last step in building up this sentence is to take 'A⊃B' and '(CvA)⊃B' and to make the first the antecedent and the second the consequent of a conditional.
Here is another example:

~{[(A≡B~)≡C]≡[C⊃(~A≡B)]}

This is a negated biconditional, and the occurrence of '≡' to which you have to apply the rule is the third. In building the sentence up from its parts, the very last thing that was done was to apply the outermost negation sign. The step before that was to form a biconditional from the components '(A≡~B)≡C' and 'C⊃(~A≡B)'. So the rule for negated biconditional applies, using '(A≡~B)≡C' and 'C⊃(~A≡B)' as the components. At the bottom of every open branch, we write Before turning you loose on some exercises, I should mention a small side point. When you worked problem Wg, one of the open branches displayed '~H' and 'K' but neither 'S' nor '~S'. So what counterexamples does this branch represent? What happened in this case is that making 'H' false and 'K' true is already enough to make everything on the branch true. If 'H' is false and 'K' is true, the initial sentences are true whether 'S' is true or false. But, strictly speaking, an assignment of truth values to sentence letters for a sentence must assign a truth value to each sentence letter in the sentence. So, strictly speaking, a counterexample for this problem must specify a truth value for 'S'. Thus we really should say that the assignment of truth values which you read off the open branch, 'H' false and 'K' true, is an abbreviation for the pair of counterexamples, ' ~H&K&S' and '~H&K&~S'.

However, having said this, we will record counterexamples by reading off the truth values for the sentence letters and negated sentence letters which occur on an open branch. If some sentence letters have been left out, we know that our list is an abbreviation for all the truth value assignments which result by arbitrarily assigning truth values for the neglected sentence letters.

Exercise

9-1. Determine the main connective in the following sentences:

a) A&[Bv(C⊃D)]

b) ~[(Hv~K)≡F]⊃)~Rv~F)

c) ~{[Iv~P0≡M]⊃(~Iv~G)}

d) {[F⊃(Bv~N)]⊃)~J⊃N)}⊃{Nv[F⊃~(BvJ)]}

9-2. Test the following arguments for validity. Show your trees, showing which paths are closed. Say whether the argument is valid or invalid, and if invalid give the counterexamples provided by the finished tree.

Before beginning these problems, you should review the practical guides at the end of chapter 8. Also, try to stay clear of the following pitfalls that often catch students: A tree is not completed until either all branches have closed or until all sentences have been checked. Sometimes you can see before completing a tree that there will surely be at least one counterexample. (This can happen, for example, when completing the left branch produced by an original sentence, before the right branch is complete.) But, both for safety and to make sure you get plenty of practice, please don't quit on a tree until all compound sentences have been checked.

Sometimes students try to take shortcuts, writing down the results of working on two rules at once. Often this produces mistakes, and makes it terribly hard for anyone to correct your papers. Finally, pay constant attention to the main connective. Only by correctly identifying the main connective in a compound sentence will you correctly apply the rules.

a) B b) DvF c) N⊃(D⊃P) d) (H&P)v(S&~J)
(BvC)&(BvD) K⊃~F N⊃D J⊃~(H&D)
~F⊃(DvK) N⊃P (J⊃~D)

e) (RvL)⊃(CvA) f) (~I&~D)v~D g) (G⊃B)&(~G⊃N)
R⊃~G ~[(I&~J)v(D&~J)] Bv~N
~(B&~R) Nv~B
~BvA

h) (D⊃H)⊃P i) I⊃(J⊃K) j) F&[Pv~(D⊃F)]
D⊃~(FvG) (I⊃J)⊃K Dv~(FvK)
FvH K⊃D
D⊃P

k) (T&G)v(G&~M) l) (H≡~Q)≡(H≡~M) m) Q≡M
T&~M H⊃(Qv~(~Q&M) (H≡~Q)≡(H≡~M)

n) J⊃[D⊃(BvP)] o) ~[~K&~(~A&~B)] p) (GvA)⊃(H⊃B)
DvP (A⊃K)&(K⊃B) [H⊃(H&B)]⊃K
H⊃P G⊃K

q) Dv(M⊃J) r) [(F&~B)vQ]⊃[A&(SvT)] s) F⊃(KvB)
[M⊃(M&J)]⊃(PvK) F&~(S&A) (~FvG)&(~Gv~K)
(P⊃D)&(K⊃F) SvA F⊃B
DvF B⊃(A⊃S)

t) F≡[~Jv(C&T)] u) Q≡~(A&F) v) (I&~T)⊃P
Av(C≡O) ~(MvA)⊃~H ~A⊃~T
~[F⊃(K⊃~O)] ~(Q&A)vF ~TvC
F≡(T&A) Q⊃(H⊃M) C⊃D
~P⊃[I⊃(D&A)]

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